Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 28}{x - 5} = \dfrac{-x + 2}{x - 5}$
Answer: Multiply both sides by $x - 5$ $ \dfrac{x^2 - 28}{x - 5} (x - 5) = \dfrac{-x + 2}{x - 5} (x - 5)$ $ x^2 - 28 = -x + 2$ Subtract $-x + 2$ from both sides: $ x^2 - 28 - (-x + 2) = -x + 2 - (-x + 2)$ $ x^2 - 28 + x - 2 = 0$ $ x^2 - 30 + x = 0$ Factor the expression: $ (x + 6)(x - 5) = 0$ Therefore $x = -6$ or $x = 5$ At $x = 5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 5$, it is an extraneous solution.